# Algebraic Extensions - Dummit and Foote, Propn 11 and 12 ...

Gold Member
I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Section 13.2 : Algebraic Extensions

I need some help with an aspect of Propositions 11 and 12 ... ...

Propositions 11 and 12 read as follows:

Now Proposition 11 states that the degree of ##F( \alpha )## over ##F## is equal to the degree of the minimum polynomial ... ... that is

##[ F( \alpha ) \ : \ F ] = \text{ deg } m_\alpha (x) = \text{ deg } \alpha##

... ... BUT ... ...

... ... Proposition 12 states that ... "if ##\alpha## is an element of an extension of degree ##n## over ##F##, then ##\alpha## satisfies a polynomial of degree at most ##n## over ##F## ... ... "

Doesn't Proposition 11 guarantee that the polynomial (the minimum polynomial) is actually of degree equal to ##n##???

Can someone please explain in simple terms how these statements are consistent?

Help will be appreciated ...

Peter

#### Attachments

• D&F - Proposition 11, Chapter 13 ....png
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• D&F - Proposition 12 and start of proof, Chapter 13 ....png
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fresh_42
Mentor
1. In order to prove (##\alpha ## algebraic ##\Rightarrow \, F(\alpha) /F## finite), we don't care exact degrees.
2. The authors mentioned, that ##\deg [F(\alpha):F] = \deg m_\alpha(x)##.
3. In the direction of the proof which you quoted, all we have is that ##\alpha## is algebraic over ##F##. This means it satisfies some polynomial equation of degree ##n##. This polynomial doesn't need to be minimal, irreducible nor has ##\alpha ## to be outside of ##F##. That it is of finite degree is all that counts. We simply don't bother more than that.

Math Amateur
andrewkirk
Homework Helper
Gold Member
This one is easier than your usual challenges!

The trouble arises because the ##\alpha## in the second sentence is not necessarily the same as the one in the first sentence. It might have been clearer if they'd used ##\beta## instead of ##\alpha## in the second and subsequent sentences.

Let the extension of interest be ##F(\alpha) / F## where ##\alpha## is the root of an irreducible quadratic in ##F[x]##.
Consider a ##\beta## that is in ##F##. That ##\beta## is also in the extension. But it is a root of the degree-1 ##F[x]## polynomial ##x-\beta##.

On the other hand, ##\alpha## is also in the extension, and the minimal polynomial for that has degree two.

So some elements of the extension (the ones that are not already in ##F##) have degree two, and some (those that are in ##F##) have degree one.

Math Amateur
Gold Member
Thanks fresh_42 and Andrew ... appreciate your help ...

Still thinking over what you have said ...

Peter